factorise using factor theorem
4z^3+23z^2 - 41x - 42
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Step-by-step explanation:
Answer: (z-2)(z+7)(4z+3)
Step-by-step explanation:
1) We have,
p(z) = 4z^3+23z^2-41z-42p(z)=4z
3
+23z
2
−41z−42
By inspection, z = 2 is zero of polynomial p(z) .
So, (z-2) will be a factor of p(z) by Factor theorem.
2) Again,
Making terms in form of above factor.
That is,
$$\begin{lgathered}4z^3+23z^2-41z-42\\ \\=4z^3-8z^2+8z^2+23z^2-41z-42\\ \\=4z^2(z-2)+31z^2-62z+62z-41z-42\\ \\=4z^2(z-2)+31z(z-2)+21z-42\\ \\=4z^2(z-2)+31z(z-2)+21(z-2)\\ \\=(z-2)(4z^2+31z+21)\\ \\=(z-2)(4z^2+28z+3z+21)\\ \\=(z-2)(4z(z+7)+3(z+7))\\ \\=(z-2)(z+7)(4z+3)\end{lgathered}$$
Hence, required factorization is
(z-2)(z+7)(4z+3)
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